If sin x = 1/3 and sec y = 5/3,where x and y lie between 0 and π/2, evaluate sin(x + y)???

x and y lie between 0 and pi/2; so that means all values of sinx, siny, cosx, cosy are positive. More information in sin x =1/3.. sine is opposite/hypotenuse. So there is a right angle triangle with angle x, the length of the line opposite it is 1 and the hypotenuse is 3. Use pythagoras a^2 + b^2 = c^2.. so a^2 + 1^2 = 3^2 so a^2 + 1 = 9.. so a^2 = 8 = 2 sqrt(2)..

now sec y =5/3 is the same as 1/cosy = 5/3 so cos y = 3/5... so using pythagoras the length of the opposite side is 4 (a 3 4 5 special triangle).

now sin(x+y) = sinx cosy + siny cosx

sin x we know is 1/3, cosy we know is 3/5... sin y is opp/hyp which is 4/5 and cos x is adj/hyp so 2sqrt(2)/3.. putting that all together:

now sin(x+y) = sinx cosy + siny cosx
= 1/3 * 3/5 + 4/5 * 2sqrt(2)/3
= 1/5 + 8sqrt(2)/15
= (3 + 8sqrt(2) )/15